Recall from [1]} that \(\Omega ^1(k[x_1,\dots , x_n])\) 
and \(\Omega ^1(k[X_1,\dots , X_m])\)  are free modules. By our hypothesis
that \(\lbrace  \mathrm {d}_x (X_1) ,\dots , \mathrm {d}_x (X_m)\rbrace \)  is free in \(\Omega ^1(k[x_1,\dots ,x_n])\) ,
it follows that the map
\(\Omega ^1(k[X_1,\ldots , X_m]) \longrightarrow \Omega ^1(k[x_1,\ldots , x_n]) , \qquad \sum _j f_j\,\mathrm {d}_X(X_j)\longmapsto \sum _j f_j \mathrm {d}_x(X_j)\) 
